Region $R$ is enclosed by the curves $y=x^3$ and $y=\sqrt x$. $y$ $x$ ${y=x^3}$ ${y=\sqrt x}$ $ 1$ $ 0$ $ 1$ $ R$ What is the volume of the solid generated when $R$ is rotated about the $y$ -axis? Give an exact answer in terms of $\pi$.
Solution: Let's imagine the solid is made out of many thin slices. Each slice is a cylinder with a hole in the middle, much like a washer. $y$ $x$ ${y=x^3}$ ${y=\sqrt x}$ $ 1$ $ 0$ $ 1$ Let the width of each slice be $dy$, let the radius of the washer, as a function of $y$, be $r_1(y)$, and let the radius of the hole, as a function of $y$, be $r_2(y)$. Then, the volume of each slice is $\pi[(r_1(y))^2-(r_2(y))^2]\,dy$, and we can sum the volumes of infinitely many such slices with an infinitely small width using a definite integral: $\int_a^b \pi [(r_1(y))^2-(r_2(y))^2]\,dy$ We call this the washer method. What we now need is to figure out the expressions for $r_1(y)$ and $r_2(y)$ and the interval of integration. $r_1(y)$ is equal to the distance from curve $y=x^3$ to the $y$ -axis. To find it, we need to solve the equation for $x$ : $x=\sqrt[3]{y}$ So, $r_1(y)=\sqrt[3]{y}}$. $r_2(y)$ is equal to the distance from the curve $y=\sqrt x$ to the $y$ -axis. To find it, we need to solve the equation for $x$ : $x=y^2$ So, ${r_2(y)=y^2}$. Now we can find an expression for the area of the washer's base: $\begin{aligned} &\phantom{=} \pi [(r_1(y)})^2-({r_2(y)})^2] \\\\ &= \pi [({\sqrt[3]{y}})^2-({y^2})^2] \\\\ &=\pi\left(y^{^{\frac23}}-y^4\right) \end{aligned}$ The bottom endpoint of $R$ is at $y=0$ and the top endpoint is at $y=1$. So the interval of integration is $[0,1]$. Now we can express the definite integral in its entirety! $\int_0^1 \left[\pi\left(y^{^{\frac23}}-y^4\right)\right]dy$ Let's evaluate the integral. $\int_0^1 \left[\pi\left(y^{^{\frac23}}-y^4\right)\right]dy=\dfrac{2\pi}{5}$ In conclusion, the volume of the solid is $\dfrac{2\pi}{5}$.